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Work done by kinetic frictional force

  1. Dec 18, 2007 #1
    1. The problem statement, all variables and given/known data
    Answer
    Chapter 6, Problem 73 Cutnell & Johnson 7th edition
    A 356 N force is pulling an 90.2-kg refrigerator across a horizontal surface. The force acts at an angle of 25.6 ° above the surface. The coefficient of kinetic friction is 0.258, and the refrigerator moves a distance of 9.55 m. Find (a) the work done by the pulling force, and (b) the work done by the kinetic frictional force.
    (a)
    Number
    3066.050223272106 Units J
    (b)
    Number
    -1798.985720775539 Units J

    --------------------------------------------------------------------------------
    First : sig figs are turned off in my course. The listed answers have to be within 2%. Please no lectures about sig fig. In any other course I obey the rules. ;)

    Incorrect.

    (a) Number 3066.050223 Units J (is correct)

    (b) Number -1964.181362 Units J (is incorrect)

    The 3066.050223J work for the part (a) is correct. Part (b) is the incorrect part I need help with.
    Wf = Ff*d*cos 25.6 where Ff = μ*m*g
    so 0.258*90.2*-9.8*9.55*cos 180 = -2178 J
    0.258*90.2*-9.8*9.55*cos 25.6 = -1964.181362
    I've exceeded the 3 tries and can see the correct answer of -1798.985720775539, but have no idea how that was arrived at. The hint given with the problem says to set ΣFy = 0 before considering the Wf . ???????
    What am I doing wrong?






    2. Relevant equations
    W= F*D cos 25.6
    Wf = Ff*d*cos 25.6 where Ff = μ*m*g
    FN = mg = weight of the refrigerator


    3. The attempt at a solution

    W done by pulling force
    W= F*D cos 25.6
    W= 356N*9.55m*cos 25.6 = 3066.050223J

    work done by the frictional force is
    .258*90.2*9.8*9.55*cos 180 = -2178J

    0.258*90.2*-9.8*9.55*cos 25.6 = -1964.181362 is incorrect. The correct answer is -1798.985720775539. ???How?????
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Dec 18, 2007 #2
    Is there a y componet to the 356 N force as well as an x componet?
     
  4. Dec 18, 2007 #3
    the 90.2 cos 25.6 considers the y. The 9.55 m moved is the x
     
  5. Dec 18, 2007 #4

    Doc Al

    User Avatar

    Staff: Mentor

    This is incorrect:
    (1) When using W = F*d*cos(theta), theta is the angle between the force and the displacement. In this case the force (friction) is horizontal and so is the displacement.
    (2) Ff = μ*N, but N is not simply mg. You need to consider the effect of the vertical component of the applied force.
     
  6. Dec 18, 2007 #5
    The normal reaction force exerted on the refrigerator,
    FN = mg - F sin 25.6 = 90.2x9.8 - 356x0.432
    = 884 - 154 = 730 N.
    Force of friction,
    f = μ FN = 0.258 x 730
    =188.3 N
    along the horizontal surface, but opposite to displacement.

    Work done by the frictional force,
    Wf = f.S = - f.S cos 180 = -f.S
    = - 188.3 x9.55 = - 1798. J
     
  7. Dec 18, 2007 #6

    Doc Al

    User Avatar

    Staff: Mentor

    Now you've got it.
     
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